# Easy proof of trivial fusion implies normal p-complement

Theorem: Suppose G is a finite group with Sylow p-subgroup P. Then the following are equivalent:

• The set K of elements of G of order relatively prime to p (the p′-elements) form a subgroup
• If A and Ag are both subsets of P, then there is some x in CG(A) such that xg is in P

That the first implies the second is a silly trick: $(a^{-1} a^g)^{-1} = g^{-1} g^a \in P \cap K = 1$ for any $a \in P$ and $g \in K$ such that $a^g \in P$.

The second implies the first is not too hard (Frobenius normal p-complement theorem), but I'm trying to use this as a first example, and so don't want to have any prerequisites outside a very gentle undergraduate group theory course. Most of the rest of the talk is just using Sylow's theorem.

Is there a very low-tech, short, few-preliminaries proof that "absolutely no fusion" implies a normal p-complement?

I would be ok with assuming P is abelian, so that we get:

• If A and Ag are both subsets of P, then g in CG(A).

I'm also fine with assuming p = 2 so that "relatively prime to p" shortens to "odd".

I don't think using the transfer is appropriate, as it won't be used again, and the whole point of this proof is to motivate learning something else.

For a "no" answer to my question, it would be sufficient to convince me that transfer is needed (and nice if you can suggest a special case where it wouldn't be needed, other than P of order 2).

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It's not hard to see that what you are asking for in the case $P$ abelian is equivalent to Burnside's Transfer Theorem. (If $A,A^g \subseteq P$ then $P$ and $P^{g^{-1}}$ are conjugate in $C_G(A)$, so $P^{hg}=P$ with $h \in C_G(A)$ and then the hypothesis of BTT hives $hg \in C_G(P)$, so $g \in C_G(A)$ and we have your hypothesis.) So you are really just asking for a transfer-free proof of BTT. – Derek Holt Jun 18 '12 at 18:48
@Derek: Thanks! I agree that when P is abelian my version and BTT are easily equivalent, so yes I am apparently asking for a transfer-free proof of the transfer theorem. I'm going to keep looking for such a published proof, but am I right in thinking there probably hasn't been one? – Jack Schmidt Jun 18 '12 at 19:05
I have never come across such a proof! – Derek Holt Jun 18 '12 at 21:40
I can't even do the special case of a self-normalizing Sylow subgroup of order $p$ without BTT. You can show by counting that you have the approrpiate number of $p'$-elements, but you still need to show that they form a subgroup. – Derek Holt Jun 19 '12 at 11:34
@Derek: That is exactly the problem (no sylow intersections means the hypothesis is basically useless). The more extreme that is, the easier the character theory argument gets (until we get a Frobenius group with complement P and kernel O_{p'}(G)), but it's still a far cry from the unsophisticated level I'm looking for. I plan on writing a negative answer in the next few days, indicating the character theory and transfer proofs as the canonical alternatives. One book I have has a survey of 5 or 6 BTT proofs (all using transfer, but in different ways). – Jack Schmidt Jun 19 '12 at 13:30